Manual of Steel Design

WIND LOAD [EX-A] Project name: Client: Address: Project locaton:

xxx xxx xxx xxx

General Data: Total length of the building, L = Total width of the building or span of gable, B = Bayspacing or spacing of rafter = Eave height of the building, HE = Ridge height of the building, HR = Solution: Sustained wind pressur, qz = CcCICzVb2 1) Basic wind speed from BNBC, Vb = 2) Structure importance coefficient, CI = 3) Velocity -to-pressure conversion coefficient, Cc = 4) Terrain exposure category =

118 49 13 10 13

35966 14935 3962 3048 3962

mm mm mm mm mm

260 kmph 161 mph 1 (Table 6.2.9, page-6-33) 4.7E-005 (Page-6-33) A

Eexposure coefficient, Cz and sustained wind pressure, qz: C4.5 qz = ( 0-15 ft) 0.368 C6 qz = (20 ft) 0.415 C9 qz = (30 ft) 0.497 C12 qz = (40 ft) 0.565 C15 qz = (50 ft) 0.624 C18 qz = (60 ft) 0.677 C21 qz = (70 ft) 0.725 C24 qz = (80 ft) 0.769 C27 qz = (90 ft) 0.81 C30 qz = (100 ft) 0.849 C35 qz = (115 ft) 0.909 5) Gust response factor, CG : CG4.5 (0-15 ft) CG6 (20 ft) CG9 (30 ft) CG12 (40 ft) CG15 (50 ft)

ft ft ft ft ft

1.174 1.324 1.586 1.803 1.991 2.16 2.313 2.454 2.584 2.709 2.9

(Table 6.2.10, page-6-33) kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 (Table 6.2.11, page-6-36)

1.654 1.592 1.511 1.457 1.418

CG18 CG21 CG24 CG27 CG30 CG35

(60 ft) (70 ft) (80 ft) (90 ft) (100 ft) (115 ft)

1.388 1.363 1.342 1.324 1.309 1.287

Average height of the gable, h = At eave height of the gable frame, qhe = For total height or average of gable frame, q h = Gust response factor at total or average height, C Gh = 6) Internal peak pressure coefficient, C'pi =

11.5 0.795 0.915 1.289 6

Hence internal pressure or internal suction = C'piqh =

ft kN/m2 kN/m2

3.506 meter

0.25 0.229

kN/m2

7) External pressure coefficient Cpe for walls: a) For transverse wind: Lower value of B/L = Higher value of B/L =

0.1 0.65

B/L =

0.42

Cpe = Cpe =

-0.5 -0.6

Windward wall, Cpe = Leeward wall, Cpe = Side or End walls, Cpe = h/B = Lower value of h/B = Higher value of h/B =

0.8 -0.56 -0.7

(Figure 6.2.5, page-6-40) (Interpolated value)

0.23

and u

6.98

degree

0.3 0.5

For u Cpe = Cpe =

0 -0.7 -0.7

and u Cpe = Cpe =

Windward roof, Cpe = Leeward roof, Cpe =

-0.84 -0.7

10 -0.9 -0.9

degree

Normal to ridge

8) Design pressure for external forces plus internal suction, p = qzC GhCpe+C'piqh Windward wall: 0-15 ft 15~20 ft 20~30 ft 30~40 ft 40~50 ft 50~60 ft 60~70 ft 70~80 ft

p= p= p= p= p= p= p= p=

1.44 1.594 1.864 2.088 2.282 2.456 2.614 2.76

kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2

0.391 0.433 0.506 0.567 0.62 0.667 0.71 0.749

klf klf klf klf klf klf klf klf

80~90 ft 90~100 ft 100~115 ft Windward roof: Leeward roof: Leeward wall: Side or End walls:

p= p= p=

2.894 3.023 3.219

kN/m2 kN/m2 kN/m2

0.786 klf 0.821 klf 0.874 klf

p= p= p= p=

-0.762 -0.597 -0.345 -0.488

kN/m2 kN/m2 kN/m2 kN/m2

-0.207 -0.162 -0.094 -0.132

klf klf klf klf

9) Design pressure for external forces plus internal pressure, p = q zCGhCpe-C'piqh Windward wall: 0-15 ft 15~20 ft 20~30 ft 30~40 ft 40~50 ft 50~60 ft 60~70 ft 70~80 ft 80~90 ft 90~100 ft 100~115 ft Windward roof: Leeward roof: Leeward wall: Side or End walls:

p= p= p= p= p= p= p= p= p= p= p=

0.982 1.136 1.406 1.63 1.824 1.998 2.156 2.302 2.436 2.565 2.761

kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2

0.267 0.308 0.382 0.443 0.495 0.542 0.585 0.625 0.661 0.696 0.75

klf klf klf klf klf klf klf klf klf klf klf

p= p= p= p=

-1.22 -1.055 -0.803 -0.946

kN/m2 kN/m2 kN/m2 kN/m2

-0.331 -0.286 -0.218 -0.257

klf klf klf klf

35.966 14.935 3.962 3.048 3.962

meter meter meter meter meter

Interpolation At eave At h 0.79519 0.914676 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.28865 0 0 0 0

0 0 0 0 0 0

degree

-0.84 -0.84

WIND LOAD [EX-B] Project name: Client: Address: Project locaton:

xxx xxx xxx xxx

General Data: Total length of the building, L = Total width of the building or span of gable, B = Bayspacing or spacing of rafter = Eave height of the building, HE =

160 65 20 20 24

Ridge height of the building, HR = Solution: Sustained wind pressur, qz = CcCICzVb2 1) Basic wind speed from BNBC, Vb = 2) Structure importance coefficient, CI = 3) Velocity -to-pressure conversion coefficient, Cc = 4) Terrain exposure category =

ft ft ft ft ft

mm mm mm mm mm

210 kmph 130 mph 1 (Table 6.2.9, page-6-33) 4.7E-005 (Page-6-33) B

Eexposure coefficient, Cz and sustained wind pressure, qz:

(Table 6.2.10, page-6-33)

C4.5

( 0-15 ft)

0.801

qz =

1.667

kN/m2

C6

(20 ft)

0.866

qz =

1.803

kN/m2

C9

(30 ft)

0.972

qz =

2.023

kN/m2

C12

(40 ft)

1.055

qz =

2.196

kN/m2

C15

(50 ft)

1.125

qz =

2.342

kN/m2

C18

(60 ft)

1.185

qz =

2.467

kN/m2

5) Gust response factor, CG : CG4.5 (0-15 ft) CG6 (20 ft) CG9 (30 ft) CG12 (40 ft) CG15 (50 ft) CG18 (60 ft)

48768 19812 6096 6096 7315

(Table 6.2.11, page-6-36) 1.321 1.294 1.258 1.233 1.215 1.201

Average height of the gable, h = At eave height of the gable frame, qhe = For total height or average of gable frame, q h = Gust response factor at total or average height, C Gh = 6) Internal peak pressure coefficient, C'pi =

22 1.81 1.855 1.286 6

Hence internal pressure or internal suction = C'piqh =

ft kN/m2 kN/m2

6.707 meter

0.25 0.464

kN/m2

7) External pressure coefficient Cpe for walls: a) For transverse wind: Lower value of B/L = Higher value of B/L =

0.1 0.65

B/L =

0.41

Cpe = Cpe =

-0.5 -0.6

Windward wall, Cpe = Leeward wall, Cpe = Side or End walls, Cpe =

0.8 -0.56 -0.7

(Figure 6.2.5, page-6-40) (Interpolated value)

h/B =

0.34

and u

7.02

degree

Lower value of h/B = Higher value of h/B =

0.3 0.5

For u Cpe = Cpe =

0 -0.7 -0.7

and u Cpe = Cpe =

Windward roof, Cpe = Leeward roof, Cpe =

-0.22 -0.7

10 0.2 -0.9

degree

Normal to ridge

8) Design pressure for external forces plus internal suction, p = qzC GhCpe+C'piqh Windward wall: 0-15 ft 15~20 ft 20~30 ft 30~40 ft 40~50 ft 50~60 ft

p= p= p= p= p= p=

2.179 2.319 2.545 2.723 2.873 3.002

kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2

0.91 0.969 1.063 1.137 1.2 1.254

klf klf klf klf klf klf

Windward roof: Leeward roof: Leeward wall:

p= p= p=

-0.061 -1.206 -0.839

kN/m2 kN/m2 kN/m2

-0.025 klf -0.504 klf -0.35 klf

Side or End walls:

p=

-1.165

kN/m2

-0.487 klf

9) Design pressure for external forces plus internal pressure, p = q zCGhCpe-C'piqh Windward wall: 0-15 ft 15~20 ft 20~30 ft 30~40 ft 40~50 ft 50~60 ft Windward roof: Leeward roof: Leeward wall: Side or End walls:

p= p= p= p= p= p=

1.251 1.391 1.617 1.795 1.945 2.074

kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2

0.523 0.581 0.675 0.75 0.812 0.866

klf klf klf klf klf klf

p= p= p= p=

-0.989 -2.134 -1.767 -2.093

kN/m2 kN/m2 kN/m2 kN/m2

-0.413 -0.891 -0.738 -0.874

klf klf klf klf

48.768 19.812 6.096 6.096 7.315

meter meter meter meter meter

Interpolation At eave At h 0

0

0

0

1.81004 1.854847 0

0

0

0

0

0 0 0 1.285516 0 0 0

degree

-0.07 -0.84

WIND LOAD [EX-A] Date: Project name: Client: Address: Project locaton:

xxx xxx xxx xxx xxx

General Data: Total length of the building, L = Total width of the building, B = Bayspacing or spacing of frame = Hight of each floor, HFL = Eave height of the building from ground level, H E = Top height of the building from ground level, H R = Height of parapet wall, HPW = Solution: Slenderness of the Building: Sustained wind pressur, qz = CcCICzVb2 1) Basic wind speed from BNBC (page-6-32), V b = 2) Structure importance coefficient, CI = 3) Velocity -to-pressure conversion coefficient, Cc = 4) Terrain exposure category =

80 45 16 10 62 70 3

ft ft ft ft ft ft ft

24384 13716 4876 3048 18897 21336 914

mm mm mm mm mm mm mm

NON SLENDER

210 kmph 130 mph 1 (Table 6.2.9, page-6-33) 4.72E-05 (Page-6-33) A

Eexposure coefficient, Cz and sustained wind pressure, qz: C4.5 qz = ( 0-15 ft) 0.368 C6 qz = (20 ft) 0.415 C9 qz = (30 ft) 0.497 C12 qz = (40 ft) 0.565 C15 qz = (50 ft) 0.624 C18 qz = (60 ft) 0.677 C21 qz = (70 ft) 0.725 C24 qz = (80 ft) 0.769 C27 qz = (90 ft) 0.81 C30 qz = (100 ft) 0.849 C35 qz = (115 ft) 0.909

(Table 6.2.10, page-6-33) 0.766 kN/m2 0.864 kN/m2 1.035 kN/m2 1.176 kN/m2 1.299 kN/m2 1.409 kN/m2 1.509 kN/m2 1.601 kN/m2 1.686 kN/m2 1.767 kN/m2 1.892 kN/m2

C40 C45 C50

(130 ft) (145 ft) (160 ft)

5) Gust response factor, CG : CG4.5 (0-15 ft) CG6 (20 ft) CG9 (30 ft) CG12 (40 ft) CG15 (50 ft) CG18 (60 ft) CG21 (70 ft) CG24 (80 ft) CG27 (90 ft) CG30 (100 ft) CG35 (115 ft) CG40 (130 ft) CG45 (145 ft) CG50 (160 ft)

qz = qz = qz =

0.965 1.017 1.065

(Table 6.2.11, page-6-36) 1.654 1.592 1.511 1.457 1.418 1.388 1.363 1.342 1.324 1.309 1.287 1.268 1.252 1.238

Mean roof level/top of parapet whichever greater, h = At eave height of the building, qHe = At mean roof level/top of parapet of the building, q h = Gust response factor at: h, CGh = h/L = Lower value of h/B = Higher value of h/B =

2.009 kN/m2 2.117 kN/m2 2.217 kN/m2

66 1.439 1.48 1.37

ft 20.122 meter kN/m2 kN/m2 (Interpolated value)

0.83

and B/L =

0.56

0.5 10

For B/L = Cpe = Cpe =

0.5 1.45 1.85

and B/L 0.65 Cpe = 1.55 Cpe = 2

1.5

(Interpolated value)

Windward wall, Cpe =

8) Design ovarall wind pressure perpendicular to wall, p = qzC GhCpe 0-15 ft 15~20 ft 20~30 ft 30~40 ft 40~50 ft 50~60 ft 60~70 ft 70~80 ft

p= p= p= p= p= p= p= p=

1.574 1.776 2.127 2.417 2.669 2.895 3.101 3.29

kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2

32.87 37.09 44.42 50.48 55.74 60.46 64.77 68.71

psf psf psf psf psf psf psf psf

F= F= F= F= F= F= F= F=

5.26 5.935 7.108 8.077 8.919 9.674 10.363 10.994

kips kips kips kips kips kips kips kips

80~90 ft 90~100 ft 100~115 ft 115~130 ft 130~145 ft 145~160 ft

p= p= p= p= p= p=

3.465 3.631 3.888 4.128 4.35 4.556

kN/m2 kN/m2 kN/m2 kN/m2 kN/m2 kN/m2

72.37 75.83 81.2 86.21 90.85 95.15

psf psf psf psf psf psf

F= F= F= F= F= F=

11.579 12.134 12.993 13.795 14.537 15.225

kips kips kips kips kips kips

24.384 13.716 4.876

meter meter meter

18.897 21.336 0.914

meter meter mm

Interpolation At eave At h 0 0 0 0 0 0 0 0 0 0 0 0 1.4389 1.479733 0 0 0 0 0 0 0 0

0 0 0

0 0 0 0 0 0 0 0 0 1.370317 0 0 0 0 0 0 0

1.49 1.91

EARTH QUAKE LOAD Height of the building, H = Height of each story, h = Number of frames of equal rigidity, NF =

100 10 4

ft ft no.

Total story of the building, n = Beam (Top Floor i.e. below roof) Serial Length Total Dimension No. (ft) No. Depth (in) Width (in) 1 10 4 14 10 2 12 6 16 10 3 14 5 18 12 4 16 5 18 6 20 Total roof slab area, A (sft)= 5000 Total length of 5 in brick wall (ft)= 100 Total length of 10 in brick wall (ft) = 0 Ceramic tiles on morter bed (per sft) 0 Suspended celling (per sft) = 10

10 no. Column (Top Floor i.e. above roof) Length Total Dimension (ft) No. Depth (in) Width (in) 3 3 3 3 3 3 Roof slab thickness, tR (in)= 4 Height of the 5 in wall (ft)= 3 Height of the 10 in wall (ft)= 3 3" Lime concrete (per sft) = 30 13 mm Celling (per sft) = 6

Beam (Typical Intermediate Floor) Serial Length Total Dimension No. (ft) No. Depth (in) Width (in) 1 10 3 14 10 2 12 6 16 12 3 14 5 18 12 4 16 5 18 6 20 Total floor slab area, A (sft)= 5000 Total length of 5 in brick wall (ft)= 120 Total length of 10 in brick wall (ft) = 80 Ceramic tiles on morter bed (per sft) 22 Suspended celling (per sft) = 10

Column (Typical Intermediate Floor) Length Total Dimension (ft) No. Depth (in) Width (in) 10 6 10 10 10 4 12 12 10 8 16 16 10 10 10 Roof slab thickness, tR (in)= 5 Height of the 5 in wall (ft)= 10 Height of the 10 in wall (ft)= 10 20 mm Floor finish (per sft) = 10 13 mm Celling (per sft) = 6

Seismic zone coefficient, Z = Structure importance coefficient, I = Response modification coefficient for structural systems, R =

0.15 1 5

Site coefficient for soil characteristics, S = Ct =

1.5 0.073

Fundamental period of vibration in seconds, T = C t.H3/4 = Numerical coefficient, C = 1.25S / T2/3 = Total seismic dead load, W = Hence, Design base shear, V = ZICW / R = Concentrated lateral force at top of the building, Ft = 0.07TV or 0.0 Distribution of Base Shear: Story

wx

hx

wxhx

No 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

(Kips) 0 0 0 0 0 503.83 746.55 746.55 746.55 746.55 746.55 746.55 746.55 746.55 746.55

(ft) 0 0 0 0 0 100 90 80 70 60 50 40 30 20 10

(Kip-ft) 0 0 0 0 0 50383 67189.5 59724 52258.5 44793 37327.5 29862 22396.5 14931 7465.5

W = 7222.8

∑ = 386330.5

0.95 seconds 1.94 7222.78 Kips 420.37 Kips 27.95 Kips (Force per frame)

wxhx/∑wihi 0 0 0 0 0 0.13 0.174 0.155 0.135 0.116 0.097 0.077 0.058 0.039 0.019 1

Fx = (V-Ft)* wxhx/∑wihi (Kips) 0 0 0 0 0 51.01 68.28 60.83 52.98 45.52 38.06 30.22 22.76 15.3 7.46 392.42

P =Fx / NF (Kips) 0 0 0 0 0 12.75 17.07 15.21 13.25 11.38 9.52 7.56 5.69 3.83 1.87

Width (in) 5250 10800 14175 0 0 0

0 0 0 0 0 0

3937.5 12960 14175 0 0 0

5625 5400 19200 0 0 0

503.83

Width (in)

746.55

PURLIN DESIGN Project name: Client: Address: Project locaton:

xxx xxx xxx xxx INPUT

Yield stress of steel, Fy =

50.041 ksi 29000 ksi

Elastic modulus, E = Bay length I.e. spacing of rafter, LBAY

CALCULATION:

34.5 Kn/cm2 19993.79 Kn/cm2

19.685 ft

6000

mm

Slope of the roof i.e pitch =

3.937 ft 5.71 degree

1200 5.71

mm degree

Design wind pressure on wind ward roof, P w =

-1.87

kN/m2

11.9 4.35 3.89

psf kg/m2 kg/m

Spacing of purlin i.e. panel length, LPANEL

IMPOSED LOAD Live load, LL = Weigth of roof sheeting, WR = Purlin mass per unit length, WLMP =

Z20016

SOLUTION Panel area supported by on purlin, APANEL = (LBAY x LPANEL) = LIVE LOAD: Total live load on each panel, WLL = APANEL x LL = Uniformly distributed live load, wLL = WLL/LBAY = DEAD LOAD: Roof deck load supported by one purlin, WP = APANEL x WR = Weight of each purlin, PP = (WLMP x LBAY) = Total dead load on each panel, WDL = (PP + WP) = Uniformly distributed dead load, wDL = WDL/LBAY = WIND LOAD: Total wind load on each panel, WWL = APANEL x Pw = Uniformly distributed wind load, wWL = WWL/LBAY = DESIGN LOAD COMBINATION:

77.5002 sft 922.25 lb 46.85 plf 68.98 51.48 120.46 6.12

lb lb lb plf

-3031.57 lb -154 plf

-39.11695 psf

0.89 2.615

psf plf

WIND LOAD ON WIND WARD ROOF: INPUT >

0.77

k / ft

OUTPUT >

11.24 1.87

Kn / m Kn / m2 0.03912 0.03912

(-)ve Sign indicates the Wind is Suction. (+)ve Sign indicates the Wind is pressure.

k/ft2 k/ft2

Chosen Wind Load Check with Above V

Uniformly distributed service load, w = wDL + wLL =

52.97 plf

0.773

KN/m

Load component perpendicular to the roof, w y = wcosu =

52.71 plf

0.769

KN/m

Load component parallel to the roof, w x = wsinu =

5.27

0.077

KN/m

2553.15 ft-lb

3.461

KN-m

255.27 ft-lb

0.346

KN-m

-147.88 plf -147.91 plf

-2.158 -2.158

KN/m KN/m

0.009

KN/m

7164.4 ft-lb

9.713

KN-m

29.55 ft-lb

0.040

KN-m

Mx = My =

2 0.1250 wyL = 2 0.1250 wxL =

Uniformly distributed load, w = wDL + wWL = Load component perpendicular to the roof, w y = wDLcosu + wWL = Load component parallel to the roof, w x = wDLsinu + 0 = Mx = My = Section

0.61

2 0.1250 wyL = 2 0.1250 wxL =

plf

plf

Z20016 whose: Sx =

35.69

x103mm3

2.18

in3

whose: Sy =

8.047

x103mm3

0.49

in3

whose: Ix =

3.48

x106mm4

8.36

in4

whose: Iy =

0.397

x106mm4

0.95

in4

Check stress, fb = Mx/Sx+My/Sy =

Check stress Ratio, [Actual Stress / Allowable Stress]

Moment Calculation for two point load for simple beam

M=

P= a=

Pa

=

55 20

1100

1) Red ink for input data 2) Magenta for Analysis data 3) Blue for AISC manual 4) Black is calculated data

X1

Y1

bf/2

X2 Y

h

tf

d tw X

bf 4000 216000

lb in-lb

110.31

mm for compactness

5.44 452.55 2.76

mm for compactness mm for compactness mm for compactness

X1 = X2 = Y1 = Y=

400 350 1500 6000

mm mm mm mm

X1 = X= Y1 = Y=

400 200 1500 6000

mm mm mm mm

X=

200

mm

X2 =

350

mm

Maximum Limit 163 mm (95/ Fy ) 557

9.42727273

mm (760/ Fy ) + 2tf

65/ Fy = 9.19239 640/ Fy = 90.51

Calculation Of Allowable Shear Strees General Data:

3.53 3.2 3.2

ft ft

a=

237.6 in

h=

9.36 in

a/h=

ft

>

2

ft ( = Lb)

25.3846

tw=

0.2 in

Calculation of kv kv=

kc =

4.01

1 5.35

(In cell E11) use kv=

5.35

h/tw=

46.8

(In cell E11)

Calculation of Cv 2.02

ues moment, M2 =

ft 38

56250 kv/Fy=

6014.48

ft-kips Cv=

2.3

2.20

NOTE: AISC ASD 9TH ED. P-(5-47) 1.33

5.708

ft use Cv=

12.764

1.33

ft

Calculation of Fv 380/Sqrt(Fy)=

53.7401

Fv=

20 22.97 Use Fv=

psi

20.00

(In cell E11)

20.00

ksi

(For plastered constructiion) (For unplastered floor constructiion) (For unplastered roof constructiion)

Defflection for Concentreted Load: 0.91

in

0.91

in

1.06 in > 0.91 in (Deflection exceeds the limit, select a beam having greater I)

Allowable Shear Strees: INPUT a= maxm. Clr./ distance between stiffeners. h= clr. Distance between two flange

tw=thck. of web

a/h1

Cv=56250kv/Fy

Cv>0.8

h/tw

Fy

EQN-2

COLUMN DESIGN INPUT DATA:

Elastic modulus, E = Yield stress of steel, Fy =

29000 50 6.64 259 25.12 5

Axial compressive force, P = Moment at end, M = Length of the column, L = No. of brace point, n =

ksi ksi kip ft-kip ft

Solution: Take effective length factor (according to support condition), K = bf =

200

mm

7.874

in

h = (d-2*tf) =

tf =

10

mm

0.394

in

d= tw =

700

mm

27.559

in

X-area, A = Ix=

5

mm

0.197

in

KLx/rx =

21.38

Iy = rx = Ix/A = ry = Iy/A =

KLy/ry =

144.4 Control

Sx =

Cc = 2p2E/Fy = Since value of Cc is Axial stress, fa = P/A = Allowable axial stress, Fa =

Sy = 107 < 0.58 7.16

Weight = than the maximum slenderness ratio ksi ksi

Allowable bending stress for maximum section modulus, Bending stress, fb = M/S = 29.36 ksi If fa/Fa is < than 0.15, check the following equation Check: fa/Fa + fb/Fb =

fa / F a = Fb = 0.60Fy =

If fa/Fa is > than 0.15, check the following two equations Euler buckling stress, F'e = 326.69 ksi

Cm =

Check (1): fa/Fa + (Cmfb)/(1-fa/F'e)Fb = Check (2): fa/0.6Fy + fb/Fb = Check of Local Stability: Check (1): bf/2tf = Check (2): h/tw = SOLUTION FOR STRONG AXIS BENDING: Take effective length factor (according to support condition), K = bf =

200

mm

7.87

in

h = (d-2*tf) =

tf =

10

mm

0.39

in

d= tw =

700

mm

27.56

in

X-area, A = Ix=

5

mm

0.2

in

KLx/rx =

Iy = rx = Ix/A = ry = Iy/A =

21.38

Sx = Sy = Cc = 2p2E/Fy =

Since value of Cc is Axial stress, fa = P/A = Allowable axial stress, Fa =

107

>

Weight =

than the maximum slenderness ratio

0.58 ksi 28.15 ksi

Allowable bending stress for maximum section modulus, Bending stress, fb = M/Sx = 29.36 ksi If fa/Fa is less than 0.15, check the following equation Check: fa/Fa + fb/Fb =

fa / F a = Fb = 0.66Fy =

If fa/Fa is more than 0.15, check the following two equations Euler buckling stress, F'e = 326.69 ksi Cm = Check (1): fa/Fa + (Cmfb)/(1-fa/F'e)Fb = Check (2): fa/0.6Fy + fb/Fb = Check of Local Stability: Check (1): bf/2tf = Check (2): h/tw =

AXIAL TENSION AND BENDING INPUT DATA: Length of the column, L = Axial compressive force, P = Moment at end, M = Elastic modulus, E = Yield stress of steel, Fy =

16 20 55 29000 36 0

No. of brace point, n =

ft kip ft-kip ksi ksi

Solution: Take effective length factor (according to support condition), K = bf =

150

mm

5.91

in

h = (d-2*tf) =

tf =

10

mm

0.39

in

d= tw =

400

mm

15.75

in

X-area, A = Ix=

5

mm

0.2

in

KLx/rx =

29.18

Iy = rx = Ix/A = ry = Iy/A =

KLy/ry =

144.36 Control

Sx = Sy =

Cc = 2p2E/Fy =

126.1

Weight =

Since value of Cc is Axial stress, fa = P/A = Allowable axial stress, Fa =

< 2.63 7.17

than the maximum slenderness ratio ksi ksi

Allowable bending stress for maximum section modulus, Bending stress, fb = M/S = 15.79 ksi If fa/Fa is less than 0.15, check the following equation Check: fa/Fa + fb/Fb = Check of Local Stability: Check (1): bf/2tf = Check (2): h/tw =

fa / F a =

Fb = 0.60Fy =

COLUMN DESIGN

2016.07607 24192.9128 48385.8257

X1

40693.3 81386.6 216.546

Y1 X2 Y

0.8 680

mm

26.77 in

7400

mm2

11.47 sq.in

607146667 mm4

1458.68 in4

13340416.7 mm4 286.44 mm 42.46 mm

32.05 in4 11.28 in 1.67 in

X1 = X2 =

400 mm 350 mm

X1 = X=

400 200

X

1734704

mm3

105.86 in3

Y1 = 1500 mm

Y1 =

1500

38115 58.08 17.71

mm3 kg/m kg/ft

2.33 in3 39.04 plf

Y=

6000 mm

Y=

6000

X=

200 mm

X2 =

350

144.4

Elastic buckling controls

0.08 30

<

0.15

10.7

ksi

14.29 24.99

1.06

>

1

BAD

EQN. H 1 - 3

0.85

(For side sway)

0.91

<

1

OK

EQN. H 1 - 1

1

<

1

BAD

EQN. H 1 - 2

9.99 135.89

< >

13.44 OK 35.78 BAD

(95/ Fy) (253/ Fy)

NOTE: EQN 1-1, 1-2 & 1-3 ARE ONLY FOR COLUMN SUBJ. TO AXIAL COMPRESSION + BENDING.

0.8 680

mm

26.77 in

7400

mm2

11.47 sq.in

ft /Ft + fb/Fb =

607146667 mm4

1458.68 in4

13340416.7 mm4 286.44 mm 42.46 mm

32.05 in4 11.28 in 1.67 in

Ft fa

= =

0.6 * Fy (ALLOWABLE TENSILE STRE AXIAL TENSILE STRESS

105.86 in3

Fb

=

ALLOWABLE BENDING STRESS

fb

=

COMPUTED AXIAL BENDING STRESS

1734704

mm3

38115

mm3

2.33

58.08 17.71

kg/m kg/ft

39.04 plf

21.38

Inelastic buckling predominates

0.02 33

0.91

<

0.15

<

1

in3

ksi

OK

EQN. H 2 - 1

0.85

(For side sway)

0.78 0.91

< <

10.09 133.85

< >

1 1

OK OK

13.44 OK 35.78 BAD

(95/ Fy) (253/ Fy)

X1

Y1 X2 Y

1 380

mm

4900

mm2

14.96 in 7.6

sq.in X

136963333 mm4

329.06 in4

5628958.33 mm4 167.19 mm 33.89 mm

13.52 in4 6.58 in 1.33 in

X1 = X2 =

400 mm 350 mm

X1 = X=

400 200

684816

mm3

41.79 in3

Y1 = 1500 mm

Y1 =

1500

28144

mm3

1.72

Y=

6000 mm

Y=

6000

38.46 11.73

kg/m kg/ft

25.85 plf X=

200 mm

X2 =

350

in3

144.36

Elastic buckling controls

0.37

>

0.15

1.1

>

1

7.58 74.8

< >

21.6

ksi

BAD

15.83 OK 42.17 BAD

(95/ Fy) (253/ Fy)

18547.2

148320

46.1

922.7813

1.5625 5.126563

mm mm mm

173.3118

mm 19.3579

mm

12.495

562.275

7.3 23.9513 285.9785 4003.699 22.967

344 272

379 275

159.8503

466

466

4635.659

422.5

496

905 199 157 300 254 251

960

3570.5

3875

WABLE TENSILE STRESS) E STRESS

BENDING STRESS

XIAL BENDING STRESS

215 168 340 293 283

mm mm mm mm mm

T ~ COLUMN DESIGN (TAPPERED COLUMN DESIGN) INPUT DATA: Elastic modulus, E = Yield stress of steel, Fy =

29000 50 116 0 26 0

Axial compressive force, P = Moment at end, M = Length of the column, L = No. of brace point, n =

ksi ksi kip ft-kip ft

Solution: Take effective length factor (according to support condition), K = bf =

325

mm

12.8

in

h = (d-2*tf) =

tf =

12

mm

0.47

in

d= tw =

300

mm

11.81

in

X-area, A = Ix=

6

mm

0.24

in

KLx/rx =

58.76

Iy = rx = Ix/A = ry = Iy/A =

KLy/ry =

93.13 Control

Sx =

Cc = 2p E/Fy = 2

Since value of Cc is Axial stress, fa = P/A = Allowable axial stress, Fa =

Sy = 107

>

Weight =

than the maximum slenderness ratio

7.91 ksi 16.26 ksi

Allowable bending stress for maximum section modulus, Bending stress, fb = M/S = 0 ksi If fa/Fa is less than 0.15, check the following equation

fa / F a = Fb = 0.60Fy =

Check: fa/Fa + fb/Fb = If fa/Fa is more than 0.15, check the following two equations Euler buckling stress, F'e = 43.25 ksi

Cm =

Check (1): fa/Fa + (Cmfb)/(1-fa/F'e)Fb = Check (2): fa/0.6Fy + fb/Fb = Check of Local Stability: Check (1): bf/2tf = Check (2): h/tw = SOLUTION FOR STRONG AXIS BENDING: Take effective length factor (according to support condition), K = bf =

150

mm

5.91

in

h = (d-2*tf) =

tf =

10

mm

0.39

in

d= tw =

400

mm

15.75

in

X-area, A = Ix=

5

mm

0.2

in

KLx/rx =

Iy = rx = Ix/A = ry = Iy/A =

47.42

Sx = Sy = Cc = 2p2E/Fy =

Since value of Cc is Axial stress, fa = P/A = Allowable axial stress, Fa =

107

>

Weight =

than the maximum slenderness ratio

15.26 ksi 24.75 ksi

Allowable bending stress for maximum section modulus, Bending stress, fb = M/Sx = 0 ksi If fa/Fa is less than 0.15, check the following equation

fa / F a = Fb = 0.66Fy =

Check: fa/Fa + fb/Fb = If fa/Fa is more than 0.15, check the following two equations Euler buckling stress, F'e = 66.41 ksi

Cm =

Check (1): fa/Fa + (Cmfb)/(1-fa/F'e)Fb = Check (2): fa/0.6Fy + fb/Fb = Check of Local Stability: Check (1): bf/2tf = Check (2): h/tw =

AXIAL TENSION AND BENDING INPUT DATA: Length of the column, L = Axial compressive force, P = Moment at end, M = Elastic modulus, E = Yield stress of steel, Fy =

16 20 55 29000 36 0

No. of brace point, n =

ft kip ft-kip ksi ksi

Solution: Take effective length factor (according to support condition), K = bf =

150

mm

5.91

in

h = (d-2*tf) =

tf =

10

mm

0.39

in

d= tw =

400

mm

15.75

in

X-area, A = Ix=

5

mm

0.2

in

KLx/rx =

29.18

Iy = rx = Ix/A = ry = Iy/A =

KLy/ry =

144.36 Control

Sx = Sy =

Cc = 2p2E/Fy =

126.1

Weight =

Since value of Cc is Axial stress, fa = P/A = Allowable axial stress, Fa =

< 2.63 7.17

than the maximum slenderness ratio ksi ksi

Allowable bending stress for maximum section modulus, Bending stress, fb = M/S = 15.79 ksi If fa/Fa is less than 0.15, check the following equation Check: fa/Fa + fb/Fb = Check of Local Stability: Check (1): bf/2tf = Check (2): h/tw =

fa / F a =

Fb = 0.60Fy =

GN

X1

Y1 X2 Y

1 276

mm

10.87 in

9456

mm2

14.66 sq.in

172346688 mm4

414.06 in4

68661218 mm4 135 mm 85.21 mm

164.96 in4 5.31 in 3.35 in

X1 =

400

mm

X1 =

400

X2 =

350

mm

X=

200

1148977

mm3

70.11 in3

Y1 =

1500 mm

Y1 =

1500

457741 74.22

mm3 kg/m

27.93 in3 49.89 plf

Y=

6000 mm

Y=

6000

22.63

kg/ft

X=

200

X2 =

350

93.13

Inelastic buckling predominates

0.49 30

> ksi

0.15

X

mm

0.49 0.85

<

1

OK

(For side sway)

0.49 0.26

< <

13.62 45.29

> >

1 1

OK OK

13.44 BAD 35.78 BAD

(95/ Fy) (253/ Fy)

1 380

mm

4900

mm2

14.96 in 7.6

sq.in

136963333 mm4

329.06 in4

5628958.33 mm4 167.19 mm 33.89 mm

13.52 in4 6.58 in 1.33 in

684816

mm3

41.79 in3

28144

mm3

1.72

38.46 11.73

kg/m kg/ft

25.85 plf

47.42

Inelastic buckling predominates

0.62 33

> ksi

0.15

in3

0.62 0.85

<

1

OK

(For side sway)

0.62 0.51

< <

7.58 74.8

< >

1 1

OK OK

13.44 OK 35.78 BAD

(95/ Fy) (253/ Fy)

X1

Y1 X2 Y

1 380

mm

4900

mm2

14.96 in 7.6

sq.in X

136963333 mm4

329.06 in4

5628958.33 mm4 167.19 mm 33.89 mm

13.52 in4 6.58 in 1.33 in

X1 =

400

mm

X1 =

400

X2 =

350

mm

X=

200

684816

mm3

41.79 in3

Y1 =

1500 mm

Y1 =

1500

28144

mm3

1.72

Y=

6000 mm

Y=

6000

38.46

kg/m

25.85 plf

in3

11.73

kg/ft

144.36

Elastic buckling controls

0.37

X=

>

0.15

1.1

>

1

7.58 74.8

< >

21.6

ksi

BAD

15.83 OK 42.17 BAD

(95/ Fy) (253/ Fy)

200

mm

X2 =

350

mm mm mm mm mm

mm mm mm mm

mm

BASE PLATE DESIGN (For Axial Load Only) As per AISC ASD - 8th Edition INPUT DATA: Total axial load of column, P = Specified concrete strength, fc' = Yield stress of steel, Fy = Width of flange of column, bf =

25

kip

Depth of column, d =

3 50 225 350

ksi ksi mm mm

8.86 in 13.78 in

Take Width of base plate, Wb = and Length of base plate, Lb=

275 380

mm mm

10.83 in 14.96 in

Width of the brick wall, =

250

mm

SOLUTION: According to column size: Required width of the footing, Wf = Required length of the footing, Lf =

355 460

mm mm

10

in

13.98 in 18.11 in

Area of concrete footing, Af = Wf x Lf =

163300 mm2

253.12 sq. in

Area of chosen base plate, Ab = Wb x Lb =

104500 mm2

161.98 sq. in

Required area of base plate, A1 = (P/0.35fc')2/Af =

2.24

sq. in

Required area of base plate, A2 = P/0.7fc =

11.9

sq. in

Required minimum area of base plate, Ar =

11.9

sq. in

'

Hence, Designed area of base plate, A =

161.98 sq. in

Actual bearing stress, fp = P/A =

0.15

ksi

m = (Lb-0.95d)/2 =

0.93

in

n = (Wb-0.8bf)/2 =

1.87

in

Thickness, t1 = 2m fp/Fy =

0.1

in

Thickness, t1 = 2n fp/Fy =

0.2

in

<

161.98 sq. in

Chosen base plate size OK

Require thickness of base plate, t =

0.2

in

6

mm

Hence, Designed base plate thickness, tb =

0.36

in

10

mm

(4 mm added for weather protection)

BASE PLATE DESIGN (For Axial Load Only) As per AISC ASD - 9th Edition

NEW BASE PLAT E DES IGN

INPUT DATA: Total axial load of column, P = Specified concrete strength, fc' = Yield stress of steel, Fy = Width of flange of column, bf =

105

kip

Depth of column, d =

3 40 300 400

ksi ksi mm mm

11.81 15.75

Take Width of base plate, Wb = and Length of base plate, Lb=

500 600

mm mm

19.69 23.62

Width of the brick wall, =

250

mm

10

SOLUTION: According to column size: Required width of the Padestal, Wf = Required length of the Padestal, Lf =

700 700

mm mm

27.56 27.56

Area of concrete Padestal, A2 = Wf x Lf =

490000

mm2

759.5

Area of chosen base plate, Ab = Wb x Lb =

300000

mm2

465

Required area of base plate, A1 = (P/0.35fc')2/A2 =

13.17

sq. in

Required area of base plate, A1 = P/0.7fc' = Required area of base plate, A1 = bfd =

50.00

sq. in

Larger of 3

186.01

sq. in

Manually InPut

Required minimum area of base plate, A1 =

186.01

sq. in

<

Hence, Designed area of base plate, A =

465

sq. in

Actual bearing stress, fp = P/A =

0.23

ksi

Allowable Bearing pressure on Support, Fp = 0.35f'c (A2/A1)

1.34

0.64, take lamda = 1.0

Manually InPut

Take, C (Maxm. of m,n & n')

5.12

Thickness, t1 = 2C fp/Fy =

0.78

in

rotection) O.K.

Require thickness of base plate, t =

0.78

in

20

Hence, Designed base plate thickness, tb =

0.94

in

24

56.52

kg/pcs

Weight of Plate

=

in in in in in

in in sq. in sq. in Larger of 3 Manually InPut

465

sq. in

Chosen base plate size OK

2.1

ksi

Manually InPut

mm mm

(4 mm added for weather protection) O.K.

ANCHOR BOLT DESIGN (Only Axial & Shear Load)

Axial pullup load, T = Horizontal shear force, V = Yield strength of steel, fy = Tensile strength of steel, Ft = Crushing strength of 28 days concrete, fc' =

21 3 36 58 3

Factor of safety, FS =

3

Bolt Diameter

Threads / in. Length

Weight

Area

Kips Kips Ksi ksi Ksi

Max. Load / Bolt

399.88

Required Bolts

(mm) 16 20

(in) 0.63 0.79

(No.) 11 10

(mm) 400 500

(Kg) 0.8 1.56

(sq.mm) 151 244

(kips) 4.52 7.31

(No.) 5 3

24

0.94

8

600

2.73

340

10.19

3

30

1.18

7

900

6.15

549

16.45

2

36

1.42

6

1000

10.04

802

24.03

1

4 24

No. mm

Used no. of Anchor Bolts, N = Bolt diameter, D = Allowable bond stress, U =

Mpa

197.09

psi

(Must be less than or equal to 350 psi) ( U = 3.4 f'c/ D )

I)

27.81

in

( Ld = Asfs/U o )

ii)

29.7

in

( Ld = 0.04Abfy/ f'c )

iii) iv)

21.92 12

in in

( Ld = 0.0004dbfy )

29.7 755

in mm

Embeded length, Ld =

Provide minimum embeded length, Ld =

(Minimum embeded length)

< 900 mm OK

Considering 4 Nos. anchor bolts per column, from the above table 24mm dia anchor bolt is sufficient. But considering weather protection selected anchor bolts = 30 mm diameter

39.355 0.583 36 58 3

Kips Kips Ksi Ksi Ksi

INPUT

ONLY

EQUATION: AREA = 0.7854 (Diam - 0.9382 P)^2 Grade A = 400MpA (fu) fy = 250 MPA Grade B = 690 Mpa Tu = fu x Area Td = Tu / S.F.

fy = 400 Mpa

S.F. = 4.0 [ Joshep E. Bowles P -432]

FOUNDATION DESIGN [SQUARE / RECTANGULAR FOOTING]

INPUT DATA: Coumn size: Long side, CLS =

18

in

457

mm

Short side, CSS =

18 9 8 200 245 4 4 60

in

457 28 8

mm mm Nos.

5 100

Ksf Ib/cft

Longitudinal column bar number, # = Number of column steel rod, n = Unfactored (service) live load, LL = Unfactored (service) dead load, DL = Base of footing below final grade, H = Ultimate concrete strength, fc' = Yield stress of steel, fy = Allowable soil pressure, qa = Unit weight fill material i.e. soil, W =

Nos. Kips Kips ft Ksi Ksi

SOLUTION: Assumed total depth of footing, D = Pressure of footing, wf = D*150 =

24

in

300

psf

Pressure of soil, ws = W*(H-D) =

200

psf

Hence, Effective soil pressure, qe = qa-wf-ws =

4500

psf

Required area of the footing, A = (DL+LL)/q e = Side of the square footing, L or B = A = Hence, Selected side of the footing, L or B =

98.89 9.94 10

ft2 ft ft

Ultimate or factored load, Pu = 1.4DL+1.7LL =

683

Kips

Net upward pressure, qu = Pu/L =

6.83

Ksf

Effective depth, d = D-4.5 = Perimeter of punching area, bo = 2(CLS+d)+2(CSS+d) =

19.5

in

150

in

Punching shear force, Vu2 = Pu-qu(CLS+d)(CSS+d) =

616.3

2

Ratio of long to short side of column, Bc = CLS/CSS = Required depth for punching, d1 = Vu2/(0.85*4 fc' bo) = as =

Kips

1 19.11

in as = 40 for interior columns

40

d2 = Vu2/(0.85*(2+4/Bc) fc' bo) =

12.74

in

30 for edge columns

d3 = Vu2/(0.85*(asd/bo+2) fc' bo) =

10.61

in

20 for corner columns

Critical section location for one-way shear action: From edge of footing, LCS = (L/2-CSS/2-d) =

31.5

in

One-way shear force, Vu1 = quLLCS = 179.29 Kips Required depth for oneway shear, d 1 = Vu1/(0.85*2 fc' L) =

13.9

in

Bending moment at column edge, Mu =1/2qu(L/2-CSS/2)2L = 616.83 Kip-ft Ru = Mu/bd2 = 162.22 psi Steel ratio, r = 0.85f'c/fy[1- 1-2Ru/(0.85*0.85f'c)] = 0.00328 Taken, equivalent constant stress block depth, a =

1.06

in

Required steel area, As = Mu/(0.9fy(d-a/2)) =

7.23

in2 and a = Asfy/0.85fc'b =

Maximum steel area for balanced steel ratio, As = rbd=

7.68

in2

Minimum steel area for shrinkage, As = 0.002bD =

5.76

in2

Minimum steel area for flexure, As = (200/fy)*bd =

7.8

in2

Therefore, adopted steel area, As = Choosen bar number, # =

7.8 6

in2

Number of bar, n = Spacing, S =

18 6.71

Nos. in c/c. in both directions

Check of bearing stress: Cross-sectional area of column, A1 = CSS*CLS =

2.25

ft2

20

Area of footing, A2 = L*B = 100 Bearing strenght at base of column, N1=0.7*0.85fc'A1= 771.12 Bearing strength at top of footing, N2 = N1 A2/A1 = 5140.8 Hence, maximum adopted value of N2 = 2N1 = 1542.24 Since Pu is less than N1 and N2, bearing stress is adequate

ft2

Required minimum dowel area, Asd = 0.005A1 =

in2

1.62

(Trial value of a = d/20)

Kips Kips Kips

1.06

mm

Greater than

2N1

Collecte Data: Sp. Gravity of coarse aggregate (C.A.) =

Value: 1.85

Unit: Mix the concrete in the field by

Sp. Gravity of fine aggregate (F.A.)=

2.65

Cement =

Sp. Gravity of Cement =

3.15

F.A. =

Fineness modulus (FM) of selected F.A. =

2.4

C.A. =

Unit weignt of dry rodded C.A.= Surface moisture contains by F.A. = Surface moisture from F.A. absorbed by C.A. =

69 3 0.1

Specified minimum strength by Structural Engr. = Standard deviation (from Table 11.2, Page- 437), s =

3000 70

lb/ft3 % % psi kg/cm2

Data from Given table & graphs:

Water = Mix the concrete in the field by Cement = F.A. = C.A. = Water = So, the Density/unit wt. of the concrete

Hence Average design strength = Water/Cement ratio from the Fig: 11.3, for value H16 = Water/Cement ratio from the Table: 11.5 = Maximum size of C.A to be used from Table: 11.6 = Workability in terms of slump From Table: 11.7 = Water in lb/ft3 of concrete (From Table: 11.8) = Approximate entrapped air content (Table: 11.8) = Bulk volume of C.A. per unit volume of concrete = (From Table: 11.4)

5319 0.57 0.75 3 12.7 2 0.65

psi

373.71 kg/cm2 (According to value of F16 and 28 days curve)

in in lb/ft3 %

(According the minimum dimension & type of Construc (According to the type of Construction) (According to slump and maximum size of C.A.)

Hence, Cement content =

22.28

lb

Weight of C.A. required =

44.85

lb

Cement =

7.07 ft3

Solid volume of F.A. required =

Water =

12.7

C.A. =

17.17 ft3

Actual quantity of water to be added =

Note:

(According to maximum size of C.A. & F.M. of F.A.)

1) Unit weight of Brick ballast = 69 lb/ft3 2) Specific gravity of Brick ballast = 1.8~2.0 3) Unit weight of Stone ballast = 100 lb/ft3

24.24 ft3 Weight of F.A. =

11.38

lb

Air (%) =

4) Specific gravity of Stone ballast = 2.6~2.8 5) Unit weight of Sand (Dry to wet) = 100~120 lb/ft3 6) Specific gravity of Sand = 2.65 7) Unit weight of Cement = 90 lb/ft3 8) Specific gravity of Cement = 3.15

n the field by weight of the materials: 22.28

lb

0.248 ft3

46.87

lb

0.469 ft3

44.89

lb

0.651 ft3

11.38 lb 0.182 ft3 n the field by proportion: (By weight) (By volume) 1 1 2.1 1.89 2.01 0.51

2.63 0.73

it wt. of the concrete

125 lb/ft3

of F16 and 28 days curve)

mum dimension & type of Construction) pe of Construction) p and maximum size of C.A.)

mum size of C.A. & F.M. of F.A.)

1.25

ft3

45.5

lb

MOMENT CONNECTION (END PLATE CONNECTION) [FOR STATIC LOAD ONLY] INPUT DATA: Bending moment, M = Shear force/End reaction, R = Yield stress of steel, Fy = Depth of girder/beam, d = Thickness of the web, tw = Flange width of the girder/beam, bf = Flange thickness of the beam, tf = (bearing type), Fv = Allowable shear stress for A325 Allowable tensile stress of one bolt, Ft = Tensile strength of electrode material (E70 electrodes), F u = Shank diameter of bolts, db =

A325

152 16 50 650 6 150 8 21 44 70

kip-ft kips ksi mm mm mm mm ksi ksi ksi

0.787 in

SOLUTION: Number of bolts above and below the tension flange:

Tensile force developed in the tension flange, Ff = M/(d - tf) = Shank area of the bolt (one), Av = 3.141 x d b2/4 =

72.16 kips

0.4869 in2 Tensile force capacity per bolt, Rt = Av x Ft = 21.42 kips Hence, Required number of bolts in tensile force zone, Nb = T / Rt = 3.36881 (Symmetrical placement above and below the tension flange should be done) Try

0.787

in Diameter bolt =

4

nos.

Welding Size (SMAW): Shear force in fillet welds, Rw = T / Lw = T / {2(bf+tf) - tw)} = Required weld size, a = Rw /(0.3Fu x 0.707) =

5.91 0.4

kips/in in

Width of the Joint Plate: Width of the joint plate, W = bf + 1 =

6.906 in

Lenght of the Joint Plate: Minimum erection clearance for bolts, E = Minimum edge distance (Center of hole to edge of joint plate), Le =

1.97 1.97

Positions of bolts above tension flange, Pf = db + 0.5 =

1.969 in

Length of the joint plate, L = d + 2(s + Le) =

33.47 in

Thickness of the Joint Plate: Moment arm for bending moment of joint plate, Pe = Pf - a - (db/4) = Ca = Cb = bf / W = Area of the tension flange, Af = bf x tf = Area of web (clear of flanges), Aw = (d - 2tf) x tw = am = CaCb(Af / Aw)1/3(Pe / db)1/4 =

1.37 in 1.09 0.925 1.86 in2 5.891 in2 0.79

Bending moment acting on the joint plate, M = amFfPe/4 =

19.52 kips-in

Required joint plate thickness, tp = 6M / (0.75Fy x W) =

0.67

in in

in

ONNECTION)

INPUT NO CHANGE 25.591 in 0.236 in 5.906 in 0.315 in (From table 7.1, pp~185)

20

mm

11

mm

176

mm

50 50

mm mm

See the AISC ASD 9th EDITION Code: Page : 4-119

50

mm

851

mm

INPUT

Pf =( db+0.5), in

Ref.: See the AISC ASD 8th EDITION Code: Page : 4-111

18

Fy (ksi) 36 42 45

A325 Ca 1.13 1.11 1.1

A490 Ca 1.14 1.13 1.12

50

1.09

1.1

mm O.K.

For FuXX

=

70

ksi

0.928

Welding Size: Top Flange to End Plate Welding

=

D=

Ff 0.928(2(bf+tf)-tw)

6.37

0.40

in

3.81

0.24

in

Web to End Plate Welding

D=

0.6Fy*tw 0.928*2

0.3xteffxFuxx 16

10.11

mm

6.06

mm

TEARING CHECK / BLOCK SHEAR Combined Shear N Tear Failior of Beam web(block Shear):

Tensile Strength of steel

=

65

No. of bolt

=

3

Depth of last bolt from top

=

200

Size of Hole

=

18

Horizontal Distance of bolt from the edge

=

50

Thick ness of web

=

5

Shear At the end (Each Row)

=

11

Avr

= =

775 1.20

sq.mm sq.inch

Ant

=

205

sq.mm

=

0.32

sq.inch

R

=

33.75

Kips

CAPACITY OF EACH ROW

SAFE & PROVIDE

L-CLEAT CONNECTION 1. One Sided Connection (Single Shear Connection) Beam Section d tw

=

325

mm

12.80

in

=

5

mm

0.20

in

bf

=

150

mm

5.91

in

tf

= =

6 16

mm mm

0.24 0.63

in in

Allowable Shearing Stress Allowable Load /Bolt Nos. of Bolts Required Vertical Edge Distance lv

= = =

14.48 29.11 2.00

KN/cm2 KN Nos.

21 6.54 3

ksi Kips Nos.

=

40

mm

1.57

in

Horizontal Edge Distance lh

= = = = =

40 75 50 257.98 50.78

mm mm mm KN KN

1.57 2.95 1.97 58 11.42

in in in Kips Kips

a b c thck.

= = = =

100 100 250 6

mm mm mm mm

3.94 3.94 9.84 0.24

in in in in

ALLOWABLE LOAD, R

=

35.99

KN

8.09

Kips

Bolt Dia

db

Vertical Spacing of Bolt Horizaontal End distance of hole in Beam End distance Bearing value from Table I-F Actual Value Bearing value from Table I-F L-Cleat Section:

1.1 Check for Bearing

UNSAFE & INCREASE THICK. OF PLATE. 2) Shear Stress> Yield Strength Ultimate Strength Allowable Shear Strength Allowable Shear Strength

= = = =

34.47 44.81 17.93 13.44

KN/cm2 KN/cm2 KN/cm2 KN/cm2

50 65 20 19.5

Ksi Ksi Ksi Ksi

Net Area Actual Shearing Stress (fv)

=

11.76

Cm2

3.65

in2

=

4.16

KN

3.02

Ksi

O.K. Shearing Stress

Gross Area Actual Shearing Stress (fv)

=

15.00

Cm2

4.65

in2

=

3.26

KN

2.37

Ksi

O.K. Shearing Stress

L-CLEAT CONNECTION 1. Two Sided Connection (Double Shear Connection) Beam Section d tw

=

325

mm

12.80

in

=

5

mm

0.20

in

bf

=

150

mm

5.91

in

tf

= =

6 16

mm mm

0.24 0.63

in in

Allowable Shearing Stress Allowable Load /Bolt Nos. of Bolts Required lv

= = =

14.48 29.11 1.00

KN/cm2 KN Nos.

21 13.09 3

ksi Kips Nos.

=

40

mm

1.57

in

lh

=

40

mm

1.57

in

Bolt Dia

db

Spacing of Bolt End distance of hole in Beam End distance Bearing value from Table I-F Actual Value Bearing value from Table I-F

= = = =

75 50 257.98 50.78

mm mm KN KN

2.95 1.97 58 11.42

in in Kips Kips

a b c thck.

= = = =

100 100 250 6

mm mm mm mm

3.94 3.94 9.84 0.24

in in in in

ALLOWABLE LOAD, R

=

71.98

KN

16.18

Kips

L-Cleat Section:

1.1 Check for Bearing

SAFE & PROVIDE 2) Shear Stress> Yield Strength Ultimate Strength Allowable Shear Strength Allowable Shear Strength

= = = =

34.47 44.81 13.79 13.44

KN/cm2 KN/cm2 KN/cm2 KN/cm2

50 65 20 19.5

Ksi Ksi Ksi Ksi

Net Area Actual Shearing Stress (fv)

=

24.24

Cm2

3.76

in2

=

2.02

KN/cm2

2.93

Ksi

O.K. Shearing Stress

Gross Area Actual Shearing Stress (fv)

=

30.00

Cm2

4.65

in2

=

1.63

KN/cm2

2.37

Ksi

O.K. Shearing Stress

EAR

ksi

(65ksi for 50 Grade & 58ksi for 36 Grade)

Nos

mm

mm

0.71

in

48.93

KN

mm

mm

Kips

Ref. AISC ASD-8th Edition P-(4-11)

CAPACITY OF EACH ROW

Table I-F

Edge Alowable Loads, Kips Distance, (for 1 fastener, 1inch. thick m Lv & Lh, in Fu = 58

Manually Provided

Manually Provided

a

1

29

1.125

32.6

1.25 1.5

36.3 43.5

1.75 2 2.25

50.8 58 65.3

2.5

72.5

2.75 3

79.8 87

b

c

Tabular Value X t X n

See the AISC ASD Manual Page 4-10

LATE.

Check for Gross or Net Area 0.4 x Fy 0.3 x Fu

[On Gross Area] [On Net Area]

Shear on Net area Governs when dia of Hole > L/(6n) L/(6n)

=

13.89

dia of Hole

=

18.00

Net Area Govern

Table I-F Edge Alowable Loads, Kips Distance, (for 1 fastener, 1inch. thick m Lv & Lh, Fu = 58 in 1 29

Manually Provided

1.125

32.6

1.25 1.5

36.3 43.5

1.75 2 2.25

50.8 58 65.3

2.5

72.5

2.75

79.8

3

Manually Provided

a

87

b

c

Tabular Value X t X n

See the AISC ASD Manual Page 4-10

Check for Gross or Net Area 0.4 x Fy 0.3 x Fu

[On Gross Area] [On Net Area]

Shear on Net area Governs when dia of Hole > L/(6n) L/(6n)

=

13.89

dia of Hole

=

16.00

Net Area Govern

Alowable Loads, Kips (for 1 fastener, 1inch. thick material) Fu = 65

Fu = 70

Fu = 100

32.5

35

50

36.6

39.4

56.3

40.6 48.8

43.8 52.5

62.5 75

56.9 65 73.1

61.3 70 78.8

87.5 100 113

81.3

87.5

125

89.4 97.5

96.3 105

138 150

overns when

mm

0.55

in

mm

0.63

in

a Govern

Alowable Loads, Kips (for 1 fastener, 1inch. thick material) Fu = 65 Fu = 70 Fu = 100 32.5

35

50

36.6

39.4

56.3

40.6 48.8

43.8 52.5

62.5 75

56.9 65 73.1

61.3 70 78.8

87.5 100 113

81.3

87.5

125

89.4

96.3

138

97.5

105

150

overns when

a Govern

mm

0.55

in

mm

0.63

in

BEAM BEARING PLATE DESIGN

INPUT DATA: Reaction force, R = Allowable unit bearing pressure on wall, Fp = Width of the bearing plate (parallel to beam), C (N) = Yield stress of steel, Fy =

20 250 8 50

kips psi in ksi

Flange width of the beam, bf =

200

mm

80 11

in2 in

SOLUTION: Required plate area, A = R/Fp = Length of bearing plate parallel to wall, B = A/C

(Generally 8 in, for 10 7.87

Provide minimum Length of bearing plate, B =

11.87 in

302

Actual bearing pressure on plate, fp = R/(B*C) Cantilever projection, n = B/2-k = Allowable bearing stress on plate, Fb = 0.75Fy =

210.61 psi 4.68 in 37.5 ksi

Say, k =

Hence thickness of bearing plate, t = 3fpn2/Fb

0.61

in

Safe Bearing Pressure on Masonry and Concrete Wall: Type of Wall: Brick: i) soft ii) medium iii) hard Concrete i) hollow units ii) solid units

Pressure (psi) 150 200 300 150 260

Poured concrete walls i) 3000 psi concrete ii) 4000 psi concrete

650 850

SIGN

INPUT ONLY OUT PUT (Generally 8 in, for 10 in wall) in REF.: SEE THE AISC ASD 8TH EDITION Along with Wall Check "K" Value

PAGE: 2-47

mm

Calculate "K" Value: 1.26

16

(Generally 1 in) mm

Thickness of Flange Depth of Welding

Provide

Masonry Bearing (Fp): In the Absence of Code regulations the Following stress apply: On Sand Stone & Lime Stone On Brick in Cement Morter On the full area of Concrete Support

= = =

0.40 ksi 0.25 ksi 0.35fc' ksi

On less than the full area of the Concrete Support

=

0.35fc' A2/A1